Question

Two bodies with kinetic energies in the ratio of 3:1 are moving with equal linear momentum. The ratio of their masses is

• A. 1:4
• B. 1:3
• C. 1:2
• D. 1:1

Hint:

Remember the inverse relationship between kinetic energy and mass when momentum is constant: \( K \propto \frac{1}{m} \). So, if kinetic energy is in ratio \( K_1:K_2 \), then mass ratio \( m_1:m_2 \) will be \( K_2:K_1 \).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem asks for the ratio of the masses of two bodies, given the ratio of their kinetic energies and the fact that they have equal linear momentum.
Step 2: Key Formula or Approach:
1. Kinetic Energy (\( K \)): \( K = \frac{1}{2}mv^2 \)
2. Linear Momentum (\( p \)): \( p = mv \)
These two are related by: \( K = \frac{p^2}{2m} \). This form is convenient when momentum is constant.
Step 3: Detailed Explanation:
Let the two bodies be 1 and 2.
Given:
- Ratio of kinetic energies: \( \frac{K_1}{K_2} = \frac{3}{1} \)
- Equal linear momentum: \( p_1 = p_2 = p \)
Using the relation \( K = \frac{p^2}{2m} \):
For body 1: \( K_1 = \frac{p^2}{2m_1} \)
For body 2: \( K_2 = \frac{p^2}{2m_2} \)
Now, take the ratio of their kinetic energies:
\[ \frac{K_1}{K_2} = \frac{\frac{p^2}{2m_1}}{\frac{p^2}{2m_2}} = \frac{m_2}{m_1} \] We are given \( \frac{K_1}{K_2} = \frac{3}{1} \).
So, \( \frac{m_2}{m_1} = \frac{3}{1} \).
The question asks for the ratio of their masses (\( m_1 : m_2 \)):
\[ \frac{m_1}{m_2} = \frac{1}{3} \]
Step 4: Final Answer:
The ratio of their masses is 1:3.